\begin{answer}
    We have
    $$
    \begin{aligned}
p(y;\lambda)&= \frac{e^{-\lambda}\lambda ^y}{y!}\\
&= \frac{1}{y!}\exp (y\log \lambda - \lambda)
\end{aligned}
$$

So here
$$
\begin{aligned}
b(y) &= 1 / y!\\
\eta &= \log \lambda\\
T(y)&= t\\
a(\eta)&= e^{\eta}
\end{aligned}
$$
\end{answer}
